We can solve the expression step by step:

### The given expression is:
\[
\left( 4 \frac{1}{6} + \frac{6}{25} \right)^2 - \left( 4 \frac{1}{6} - \frac{6}{25} \right)^2
\]

This is a difference of squares, which follows the formula:
\[
a^2 - b^2 = (a - b)(a + b)
\]

Let:
\[
a = 4 \frac{1}{6} + \frac{6}{25}, \quad b = 4 \frac{1}{6} - \frac{6}{25}
\]

Thus, we can rewrite the expression as:
\[
\left( (4 \frac{1}{6} + \frac{6}{25}) - (4 \frac{1}{6} - \frac{6}{25}) \right) \times \left( (4 \frac{1}{6} + \frac{6}{25}) + (4 \frac{1}{6} - \frac{6}{25}) \right)
\]

#### Step 1: Simplifying each part

- First term: 
\[
(4 \frac{1}{6} + \frac{6}{25}) - (4 \frac{1}{6} - \frac{6}{25}) = \frac{6}{25} + \frac{6}{25} = \frac{12}{25}
\]

- Second term:
\[
(4 \frac{1}{6} + \frac{6}{25}) + (4 \frac{1}{6} - \frac{6}{25}) = 4 \frac{1}{6} + 4 \frac{1}{6} = 2 \times 4 \frac{1}{6} = 8 \frac{2}{6} = 8 \frac{1}{3}
\]

Now convert the mixed number \(8 \frac{1}{3}\) into an improper fraction:
\[
8 \frac{1}{3} = \frac{25}{3}
\]

#### Step 2: Multiply the two simplified terms:
\[
\frac{12}{25} \times \frac{25}{3}
\]

Cancel the \(25\)s:
\[
\frac{12}{3} = 4
\]

### Final answer:
\[
4
\]

Would you like more details or have any questions?

Here are some related questions to expand on the topic:
1. How do you simplify complex fractions in algebra?
2. What is the difference between a mixed number and an improper fraction?
3. Can you apply the difference of squares formula to other algebraic expressions?
4. How would you solve if the fractions involved were not rational?
5. How do you convert between mixed numbers and improper fractions?

**Tip**: The difference of squares formula is useful when you encounter expressions with two squared terms being subtracted.

Learn how to solve the algebraic expression [4⅙ + (6/25)]² - [4⅙ - (6/25)]² using the difference of squares formula. This step-by-step solution covers key concepts such as mixed numbers, fractions, and algebraic identities, suitable for students in Grades 8-10.

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